Relation Isomorphism Preserves Reflexivity
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Theorem
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be (relationally) isomorphic.
Then $\RR_1$ is a reflexive relation if and only if $\RR_2$ is also a reflexive relation.
Proof
Without loss of generality it is necessary to prove only that if $\RR_1$ is reflexive then $\RR_2$ is reflexive.
Let $\phi: S \to T$ be a relation isomorphism.
Let $y \in T$.
Let $x = \map {\phi^{-1} } y$.
As $\phi$ is a bijection it follows from Inverse Element of Bijection that:
- $y = \map \phi x$
As $\RR_1$ is a reflexive relation it follows that:
- $x \mathrel {\RR_1} x$
As $\phi$ is a relation isomorphism it follows that:
- $\map \phi x \mathrel {\RR_2} \map \phi x$
Hence the result.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.9 \ \text{(c)}$